If A is symmetric and $A^2 = A$, show it is a projection matrix.

Question

I assumed showing $A=QQ^T$ was enough to say it is a projection matrix, but I think it's not. I showed this by saying $A=QDQ^{-1}$ (D is diagonal) and knowing $Q^T = Q^{-1}$ for orthogonal matrices, we know $A^2 = A$ so we can get that $D^2 = D$. How could I proceed from here?

Answer 1

Let $V=A(\mathbb{H})$ and $W= (I-A)(\mathbb{H})$ Then $Ax=x$ for $x\in V$ and $Ax=0$ for $x\in W.$ The subspaces $V$ and $W$ are orthogonal to each other, as $$\langle Ax,(I-A)y\rangle =\langle (A-A^2)x,y\rangle =0$$ Moreover $V+W=\mathbb{H}$ as $x=Ax+(I-A)x.$ Hence $A$ is the orthogonal projection onto $V.$

Remark I have made a proof for a bounded symmetric operator on the Hilbert space $\mathbb{H},$ but we may assume that $\mathbb{H}=\mathbb{R}^n$ and $A$ is $n\times n$ symmetric matrix.